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Question: There are two matrices $A$ and $B$ of order $3\times3$ with $AB+BA=0$, then is it necessary that $AB=0$?

I tried this problem but couldn't reach any conclusive answer. I didn't see any reason for $AB=0$, then I tried to get some counterexample. I couldn't find it.

Please give me some counterexample if it exists. Also, please tell me how you reached there? I suppose it is not just a matter to attempt with two matrices by trial only. There should be some thought process to find such matrices.

In case if you find $AB=0$, please prove it.

Thank you.

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5 Answers 5

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Take the following matrices: $$ A=\begin{pmatrix} -1 & 0 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 1\end{pmatrix},\; B=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 0 & 0 \cr 1 & 0 & 0\end{pmatrix}. $$ Then $AB+BA=B-B=0$ and $$ AB=B\neq 0. $$ The idea is to take a diagonal matrix for $A$ and a nonzero matrix $B$ with $AB=B$ and $BA=-B$.

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Let $A=\begin{bmatrix}0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$, $B=diag(1,-1,0)$

$$AB=\begin{bmatrix}0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

and $$BA=\begin{bmatrix}0 & 1 & 0\\ -1 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$$

We have then found a counterexample.

I use a permutation matrix and play with the $2$ by $2$ case first before working on the $3$ by $3$ case.

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Hint: embed the Pauli-Matrices $\sigma_x$ and $\sigma_z$ in the wanted space...

EDIT: in general this is called an anti-commutative property.

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Suppose the entries of the matrices are taken from a field of characteristic $\ne2$. Let $X=A+B$ and $Y=A-B$. Then the statement $AB+BA=0$ is equivalent to $X^2=Y^2$, and the statement that $AB=0$ is equivalent to $XY=YX$.

So, to construct a counter-example, it suffices to find a pair of non-commuting matrices $X,Y$ such that $X^2=Y^2$. This is easy. E.g. Siong Thye Goh’s answer can be obtained by putting $$ X=\pmatrix{0&2&0\\ 0&0&0\\ 0&0&0}, \ Y=\pmatrix{0&0&0\\ 2&0&0\\ 0&0&0}, \ A=\pmatrix{0&1&0\\ 1&0&0\\ 0&0&0}, \ B=\pmatrix{0&1&0\\ -1&0&0\\ 0&0&0}, $$ while the answer by Dietrich Burde (up to permutations of rows and columns) may be obtained by putting $$ X=\pmatrix{1&1&0\\ 0&-1&0\\ 0&0&0}, \ Y=\pmatrix{1&-1&0\\ 0&-1&0\\ 0&0&0}, \ A=\pmatrix{1&0&0\\ 0&-1&0\\ 0&0&0}, \ B=\pmatrix{0&1&0\\ 0&0&0\\ 0&0&0}. $$ By carefully picking $X$ and $Y$, one may obtain a counter-example with a nice but rather non-trivial appearance. E.g. take two pairs of linearly independent real integer vectors $\{u,x\}$ and $\{v,y\}$ such that $\langle u,v\rangle=\langle x,y\rangle=0\ne\langle x,v\rangle$. Put $X=2uv^T$ and $Y=2xy^T$. Then $X^2=Y^2=0$ but $XY\ne YX$. Hence $A=\frac12(X+Y)=uv^T+xy^T$ and $B=\frac12(X-Y)=uv^T-xy^T$ constitute a counterexample. E.g. $$ u=\pmatrix{1\\ 1\\ 1}, \ v=\pmatrix{1\\ 2\\ -3}, \ x=\pmatrix{2\\ 3\\ 5}, \ y=\pmatrix{1\\ 1\\ -1}, \ A=\pmatrix{3&4&-5\\ 4&5&-6\\ 6&7&-8}, \ B=\pmatrix{-1&0&-1\\ -2&-1&0\\ -4&-3&2}. $$

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Hint: take $B$ non-zero that is conjugate to $-B$. Possibilities: $B$ with eigenvalues $-1$, $0$, $1$. Or $B$ a $3\times 3$ Jordan cell corresponding to $\lambda=0$. So there exists $A$ ( can be found explicitly) such that $$A B A^{-1} = -B$$

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